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Does the equation $x^3+x+4=y^2$ have positive integer solutions?

$x^3+48=y^4$ does not have integer (?) solutionsDiophantine equation-positive solutionshow many positive integer solutions to the following equation?Why does the diophantine equation $x^2+x+1=7^y$ have no integer solutions?Does the equation $x^2+23y^2=2z^2$ have integer solutions?Diophantine equation no integer solutionsInteger solutions to fraction equationDetermining number of integer solutions of Diophantine equationInteger solutions to a Diophantine EquationDoes the equation $a^2+d^2+4=b^2+c^2$ have any solutions?

5

5

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solutions?

diophantine-equations

share|cite|improve this question

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  According to the graph, only have integer solution, $(0, 2), (0,-2)$, thus this have not natural solutions for $x$.
  $endgroup$
  – Eduardo S.
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @John111: If your $mathbb{N}$ does not include $0$, you need to specify it in the question. A lot of people take $mathbb{N}$ to include $0$.
  $endgroup$
  – user10354138
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, this and this suggests there are no integer points on the curve other than $(0,pm 2)$.
  $endgroup$
  – user10354138
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance This is an elliptic curve and has infinitely many point on it, so I am not sure what do you mean by that is has no solutions?
  $endgroup$
  – Anurag A
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance but that claim is incorrect, e.g. in $y^2=x+1$, we cannot write the RHS as a square but it has infinitely many integer solutions $(k^2-1,k)$.
  $endgroup$
  – Anurag A
  9 hours ago
  

  
  
  
  

 | 
show 10 more comments

5

5

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solutions?

diophantine-equations

share|cite|improve this question

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  According to the graph, only have integer solution, $(0, 2), (0,-2)$, thus this have not natural solutions for $x$.
  $endgroup$
  – Eduardo S.
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @John111: If your $mathbb{N}$ does not include $0$, you need to specify it in the question. A lot of people take $mathbb{N}$ to include $0$.
  $endgroup$
  – user10354138
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, this and this suggests there are no integer points on the curve other than $(0,pm 2)$.
  $endgroup$
  – user10354138
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance This is an elliptic curve and has infinitely many point on it, so I am not sure what do you mean by that is has no solutions?
  $endgroup$
  – Anurag A
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @NoChance but that claim is incorrect, e.g. in $y^2=x+1$, we cannot write the RHS as a square but it has infinitely many integer solutions $(k^2-1,k)$.
  $endgroup$
  – Anurag A
  9 hours ago
  

  
  
  
  

 | 
show 10 more comments

$begingroup$

$$x^3+x+4=y^2$$

and $x,y inBbb N$ and $x,y ≠ 0$

Does this equation have solutions?

diophantine-equations

share|cite|improve this question

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 48 mins ago

YuiTo Cheng

3,38571546

edited 48 mins ago

YuiTo Cheng

3,38571546

asked 9 hours ago

John111John111

414

New contributor

John111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

John111John111

414

4 Answers
4

active

oldest

votes

10

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

15.9k21129

$endgroup$

add a comment | 

2

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2not{equiv_4} 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 9 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

2

$begingroup$

Partial solution:

Note that if $x,y in mathbb{N}$ and $x^{3} + x + 4 = y^{2}$

Then $x^{3} + x equiv y^{2} mod(4)$

$y^{2} equiv r mod(4)$ with $r in {0,1}$

So, we have that $x^{3} + x equiv 0mod(4)$ or $x^{3} + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbb{N}$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbb{N}$

and $y = 2q$ for some $q in mathbb{N}$

Thus $64k^{3} + 4k + 4 = 4q^{2}$ if and only if

$16k^{3} + k + 1 = q^{2}$

share|cite|improve this answer

answered 9 hours ago

ZAFZAF

7258

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  9 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

If you just want to know whether there are other integer solutions. You can ask a CAS for that. If you go to the online MAGMA calculator and input

Q<x> := PolynomialRing(Rationals());

E00  := EllipticCurve(x^3+x+4);

Q00  := IntegralPoints(E00);

Q00;

It will reply

[ (0 : 2 : 1), (4128 : 265222 : 1) ]

which means there are only two pairs of integer solutions.
$$(x,y) = (0,pm 2)quadtext{ and }quad( 4128,pm 265222 )$$

share|cite|improve this answer

answered 33 mins ago

achille huiachille hui

98.2k5134267

$endgroup$

add a comment | 

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10

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

15.9k21129

$endgroup$

add a comment | 

10

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

15.9k21129

$endgroup$

add a comment | 

$begingroup$

Yes, there is an integral solution with $x,yneq 0$.

The elliptic curve $y^2=x^3+x+4$ has conductor $delta=6976$. A search of John Cremona's table (Table Seven, curve 6976c1) gives integral points $x=0$ (ignore this) and $x=4128$, and a quick computation gives the corresponding $y=pm265222$.

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

15.9k21129

$endgroup$

share|cite|improve this answer

answered 8 hours ago

user10354138user10354138

15.9k21129

answered 8 hours ago

user10354138user10354138

15.9k21129

answered 8 hours ago

user10354138user10354138

15.9k21129

2

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2not{equiv_4} 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 9 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

2

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2not{equiv_4} 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 9 hours ago

AquaAqua

53.7k1365135

$endgroup$

add a comment | 

$begingroup$

Partial answer.

Write $$ x^3+x+68 = y^2+64$$
so $$(x+2)(x^2-4x+17)=y^2+8^2$$

If $x=4k+1$ then $x+2equiv 3 pmod 4$ so there exists prime $pmid x+2$ and $pequiv 3 pmod 4$ which means that $pmid y$ and $pmid 8$ so $p=2$ which is impossible.

If $x=4k+3$ then $y=2n$ so $$x^3+x+4equiv_4 2not{equiv_4} 0equiv_4 y^2$$

So we must check what happens if $x$ is even...

share|cite|improve this answer

answered 9 hours ago

AquaAqua

53.7k1365135

$endgroup$

share|cite|improve this answer

answered 9 hours ago

AquaAqua

53.7k1365135

answered 9 hours ago

AquaAqua

53.7k1365135

answered 9 hours ago

AquaAqua

53.7k1365135

2

$begingroup$

Partial solution:

Note that if $x,y in mathbb{N}$ and $x^{3} + x + 4 = y^{2}$

Then $x^{3} + x equiv y^{2} mod(4)$

$y^{2} equiv r mod(4)$ with $r in {0,1}$

So, we have that $x^{3} + x equiv 0mod(4)$ or $x^{3} + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbb{N}$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbb{N}$

and $y = 2q$ for some $q in mathbb{N}$

Thus $64k^{3} + 4k + 4 = 4q^{2}$ if and only if

$16k^{3} + k + 1 = q^{2}$

share|cite|improve this answer

answered 9 hours ago

ZAFZAF

7258

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Partial solution:

Note that if $x,y in mathbb{N}$ and $x^{3} + x + 4 = y^{2}$

Then $x^{3} + x equiv y^{2} mod(4)$

$y^{2} equiv r mod(4)$ with $r in {0,1}$

So, we have that $x^{3} + x equiv 0mod(4)$ or $x^{3} + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbb{N}$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbb{N}$

and $y = 2q$ for some $q in mathbb{N}$

Thus $64k^{3} + 4k + 4 = 4q^{2}$ if and only if

$16k^{3} + k + 1 = q^{2}$

share|cite|improve this answer

answered 9 hours ago

ZAFZAF

7258

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That concurs with my analysis, though I tried writing it $k(16k^2+1)=(q+1)(q-1)$ and if $q$ is odd then the right-hand side is divisible by $8$ - hence $k$ is divisible by $8$; and if $q$ is even then $k$ is odd.
  $endgroup$
  – Mark Bennet
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Partial solution:

Note that if $x,y in mathbb{N}$ and $x^{3} + x + 4 = y^{2}$

Then $x^{3} + x equiv y^{2} mod(4)$

$y^{2} equiv r mod(4)$ with $r in {0,1}$

So, we have that $x^{3} + x equiv 0mod(4)$ or $x^{3} + x equiv 1 mod(4)$

Then, the only solutions of this equations is $x in mathbb{N}$ such that $x equiv 0 mod(4)$

Then we have that $x = 4k$ for some $k in mathbb{N}$

and $y = 2q$ for some $q in mathbb{N}$

Thus $64k^{3} + 4k + 4 = 4q^{2}$ if and only if

$16k^{3} + k + 1 = q^{2}$

share|cite|improve this answer

answered 9 hours ago

ZAFZAF

7258

$endgroup$

share|cite|improve this answer

answered 9 hours ago

ZAFZAF

7258

answered 9 hours ago

ZAFZAF

7258

answered 9 hours ago

ZAFZAF

7258

1

$begingroup$

If you just want to know whether there are other integer solutions. You can ask a CAS for that. If you go to the online MAGMA calculator and input

Q<x> := PolynomialRing(Rationals());

E00  := EllipticCurve(x^3+x+4);

Q00  := IntegralPoints(E00);

Q00;

It will reply

[ (0 : 2 : 1), (4128 : 265222 : 1) ]

which means there are only two pairs of integer solutions.
$$(x,y) = (0,pm 2)quadtext{ and }quad( 4128,pm 265222 )$$

share|cite|improve this answer

answered 33 mins ago

achille huiachille hui

98.2k5134267

$endgroup$

add a comment | 

1

$begingroup$

If you just want to know whether there are other integer solutions. You can ask a CAS for that. If you go to the online MAGMA calculator and input

Q<x> := PolynomialRing(Rationals());

E00  := EllipticCurve(x^3+x+4);

Q00  := IntegralPoints(E00);

Q00;

It will reply

[ (0 : 2 : 1), (4128 : 265222 : 1) ]

which means there are only two pairs of integer solutions.
$$(x,y) = (0,pm 2)quadtext{ and }quad( 4128,pm 265222 )$$

share|cite|improve this answer

answered 33 mins ago

achille huiachille hui

98.2k5134267

$endgroup$

add a comment | 

$begingroup$

If you just want to know whether there are other integer solutions. You can ask a CAS for that. If you go to the online MAGMA calculator and input

Q<x> := PolynomialRing(Rationals());

E00  := EllipticCurve(x^3+x+4);

Q00  := IntegralPoints(E00);

Q00;

It will reply

[ (0 : 2 : 1), (4128 : 265222 : 1) ]

which means there are only two pairs of integer solutions.
$$(x,y) = (0,pm 2)quadtext{ and }quad( 4128,pm 265222 )$$

share|cite|improve this answer

answered 33 mins ago

achille huiachille hui

98.2k5134267

$endgroup$

Q<x> := PolynomialRing(Rationals());

E00  := EllipticCurve(x^3+x+4);

Q00  := IntegralPoints(E00);

Q00;

[ (0 : 2 : 1), (4128 : 265222 : 1) ]

share|cite|improve this answer

answered 33 mins ago

achille huiachille hui

98.2k5134267

answered 33 mins ago

achille huiachille hui

98.2k5134267

answered 33 mins ago

achille huiachille hui

98.2k5134267